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mario62 [17]
3 years ago
13

Which statement describes the reactions in an electrochemical cell

Chemistry
1 answer:
34kurt3 years ago
6 0

Answer & explanation:

Summary on electrochemical cells and redox reactions:

Electrochemical cells (or batteries) can be defined as devices capable of transforming chemical energy into electrical energy through spontaneous reactions of redox, in which electron transfer occurs.

Redox it is a chemical reaction in which there is the occurrence of oxidation and reduction of atoms of substances (chemical species) present in the process.

<u>Oxidation</u> is the loss of electrons by an atom of a chemical species, while <u>reduction</u> is the gain of electrons by an atom of a chemical species.

Thus, during an oxirreduction reaction, electrons move from the species that loses them towards the species that will receive them. This "movement" results in the formation of an electric current (or electrical energy) as occurs with batteries, for example.

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What is the electron configuration of neptunium?
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Determine the rate law and the value of k for the following reaction using the data provided.
blsea [12.9K]

<u>Answer:</u> The rate law expression is \text{Rate}=k[NO]^2[O_2]^1 and value of 'k' is 1.727\times 10^3M^{-2}s^{-1}

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

Rate law expression for the reaction:

\text{Rate}=k[NO]^a[O_2]^b

where,

a = order with respect to nitrogen monoxide

b = order with respect to oxygen

  • <u>Expression for rate law for first observation:</u>

8.55\times 10^{-3}=k(0.030)^a(0.0055)^b       ....(1)

  • <u>Expression for rate law for second observation:</u>

1.71\times 10^{-2}=k(0.030)^a(0.0110)^b       ....(2)

  • <u>Expression for rate law for third observation:</u>

3.42\times 10^{-2}=k(0.060)^a(0.0055)^b      ....(3)

Dividing 1 from 2, we get:

\frac{1.71\times 10^{-2}}{8.55\times 10^{-3}}=\frac{(0.030)^a(0.0110)^b}{(0.030)^a(0.0055)^b}\\\\2=2^b\\b=1

Dividing 1 from 3, we get:

\frac{3.42\times 10^{-2}}{8.55\times 10^{-3}}=\frac{(0.060)^a(0.0055)^b}{(0.030)^a(0.0055)^b}\\\\2^2=2^a\\a=2

Thus, the rate law becomes:

\text{Rate}=k[NO]^2[O_2]^1

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

8.55\times 10^{-3}=k[0.030]^2[0.0055]^1\\\\k=1.727\times 10^3M^{-2}s^{-1}

Hence, the rate law expression is \text{Rate}=k[NO]^2[O_2]^1 and value of 'k' is 1.727\times 10^3M^{-2}s^{-1}

4 0
3 years ago
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